Can we hopefully tip our fellow riddle solvers, by saying that I used Fermat’s little theorem in solving this myself?

]]>By: Steven Millerhttp://mathriddles.williams.edu/?p=35&cpage=1#comment-1671
Mon, 05 Dec 2011 13:54:50 +0000http://mathriddles.williams.edu/?p=35#comment-1671I don’t believe this logic is correct.
]]>By: Thilinahttp://mathriddles.williams.edu/?p=35&cpage=1#comment-1639
Mon, 05 Dec 2011 06:20:40 +0000http://mathriddles.williams.edu/?p=35#comment-1639got it in an easy way! if you notice the set of numbers, the number : 111111111 is devisible by 9 and 2007 is also devisible by 9. So that answers will be respectively, 12345679 and 223. as the set contains numbers more than 9*223, we can find a number which has 9*223, 1s: that means 2007th number. So this number is devisable, so that set contains atleast one number which is divisible by 2007.

second part : no it’s not true. it’s not divisible. if you consider the ending of 2008, is 8. when 8 is multiplied by any number from 1 to 9, the ending number is again even. but no number in the given set it even. So it’s not divisible